Most Beautiful Item for Each Query

Problem Description

You are given a 2D integer array items where items[i] = [price_i, beauty_i] denotes the price and beauty of an item respectively.

You are also given a 0-indexed integer array queries. For each queries[j], you want to determine the maximum beauty of an item whose price is less than or equal to queries[j]. If no such item exists, then the answer to this query is 0.

Return an array answer of the same length as queries where answer[j] is the answer to the j^th query.

Example 1:

Input: items = [[1,2],[3,2],[2,4],[5,6],[3,5]], queries = [1,2,3,4,5,6]
Output: [2,4,5,5,6,6]
Explanation:
- For queries[0]=1, [1,2] is the only item which has price <= 1. Hence, the answer for this query is 2.
- For queries[1]=2, the items which can be considered are [1,2] and [2,4]. 
  The maximum beauty among them is 4.
- For queries[2]=3 and queries[3]=4, the items which can be considered are [1,2], [3,2], [2,4], and [3,5].
  The maximum beauty among them is 5.
- For queries[4]=5 and queries[5]=6, all items can be considered.
  Hence, the answer for them is the maximum beauty of all items, i.e., 6.

Example 2:

Input: items = [[1,2],[1,2],[1,3],[1,4]], queries = [1]
Output: [4]
Explanation: 
The price of every item is equal to 1, so we choose the item with the maximum beauty 4. 
Note that multiple items can have the same price and/or beauty.

Example 3:

Input: items = [[10,1000]], queries = [5]
Output: [0]
Explanation:
No item has a price less than or equal to 5, so no item can be chosen.
Hence, the answer to the query is 0.

Constraints:

1 <= items.length, queries.length <= 10⁵
items[i].length == 2
1 <= price_i, beauty_i, queries[j] <= 10⁹

Solution (Go)

func maximumBeauty(items [][]int, queries []int) []int {
	sort.Slice(items, func(i, j int) bool {
		return items[i][0] < items[j][0]
	})
	beauty := 0
	cleanedItems := make([][]int, 0)
	for _, item := range items {
		if item[1] > beauty {
			beauty = item[1]
			cleanedItems = append(cleanedItems, item)
		}
	}
	res := make([]int, len(queries))
	for i, query := range queries {
		res[i] = find(cleanedItems, query)
	}
	return res
}

func find(items [][]int, query int) int {
	left, right := 0, len(items)-1
	for left <= right {
		mid := (left + right) / 2
		if items[mid][0] > query {
			right = mid - 1
		} else {
			left = mid + 1
		}
	}
	if left-1 >= 0 {
		return items[left-1][1]
	}
	return 0
}